From the voltage equation $V = 220 \sin(2 \times 10^3 t)$, we identify the peak voltage $V_0 = 220 \text{ V}$ and angular frequency $\omega = 2 \times 10^3 \text{ rad/s}$. Given components are $L = 10 \text{ mH} = 10^{-2} \text{ H}$, $C = 25 \mu\text{F} = 25 \times 10^{-6} \text{ F}$, and $R = 100 \Omega$.
Inductive reactance $X_L = \omega L = (2 \times 10^3) \times (10^{-2}) = 20 \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{(2 \times 10^3) \times (25 \times 10^{-6})} = \frac{1}{0.05} = 20 \Omega$.
Since $X_L = X_C$, the circuit is in resonance. The impedance $Z$ is equal to the resistance $R$.
$$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + (20 - 20)^2} = 100 \Omega$$
The current amplitude $I_0$ is given by:
$$I_0 = \frac{V_0}{Z} = \frac{220}{100} = 2.2 \text{ A}$$
Final Answer: 2.2 A
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